Question

崭新的蟒蛇和难倒的所有准备好，会很感激一手。

testn1 = {'names':('tn1_name1','tn1_name2','tn1_name3'),'exts':('.log','.txt')}
testn2 = {'names':('tn2_name1'),'exts':('.nfo')}
testnames = {1:testn1,2:testn1}

directory = 'C:\\temp\\root\\'

for subdir in os.listdir(directory):

  # check if name of sub directory matches the name in any of the dicts in testnames[testn*]['names']
  if os.path.isdir(os.path.join(directory, subdir)) and [subdir in subdir.lower() in testnames[testn1]['names']]: # this  works but need to iterate through all dicts
    print(subdir)

    # if the a dir name matches do a recursive search for all filenames that exist in the same dict with the corresponding extensions
    for dirname, dirnames, filenames in os.walk(os.path.join(directory, subdir)):
      for file in filenames:
        if file.endswith(testnames[testn1]['exts']): # this works but need to match with corresponding folder 
          print(file)

我以为我能够做这样的事情，但我确信我对python的理解不是它需要的。

if os.path.isdir(os.path.join(directory, subdir)) and [subdir in subdir.lower() in [for testnames[key]['names'] in key, value in testnames.items()]]:

我希望以这种方式保持结构，但对任何事情都是开放的。

编辑：我最终选择......

if os.path.isdir(os.path.join(directory, subdir)) and [i for i in testnames.values() if subdir.lower() in i['names']]:

感谢@ pzp1997关于.values（）

的抬头

Answer 1

不完全确定你想要什么，但我认为就是这样：

if os.path.isdir(os.path.join(directory, subdir)) and subdir.lower() in [i['names'] for i in testnames.values()]

Answer 2

这样做了！

if os.path.isdir(os.path.join(directory, subdir)) and [i for i in testnames.values() if subdir.lower() in i['names']]:

Answer 3

相反呢？

testn1 = {'names':('tn1_name1','tn1_name2','tn1_name3'),'exts':('.log','.txt')}
testn2 = {'names':('tn2_name1'),'exts':('.nfo')}
testnames = {1:testn1,2:testn1}

directory = 'C:\\temp\\root\\'

for dirname, _, filenames in os.walk(directory):
    the_dir =  os.path.split(dirname)[-1]
    for testn in testnames.itervalues():
        if the_dir in testn['names']:
            for file in filenames:
                _, ext = os.path.splitext(file)
                if ext in testn['exts']:
                    print the_dir, file

如果列表中的项目与字典中的子元素列表中的项目匹配

3 个答案: